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Proof step 1 from proof nat2int-05 (conclusion) (back to tree)

⊦ n→i(m) = n→i(n) + 1 ↔ m = n + 1

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof nat2int-05 (previous step) (back to tree)

¬(n→i(m) = n→i(n) + 1 ↔ m = n + 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: nat2int-03: ⊦ n→i(n + 1) = n→i(n) + 1; add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 ⊦ i + j = i + i0 ↔ j = i0; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic

Proof step 3 from proof nat2int-05 (previous step) (back to tree)

n→i(m) = n→i(n) + 1, m ≠ n + 1 ⊦

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: m

Additional information

Used simplifier rules: nat2int-07: ⊦ 0 ≠ n→i(m) + 1; nat2int-03: ⊦ n→i(n + 1) = n→i(n) + 1; inj: ⊦ n→i(m) = n→i(n) ↔ m = n; nat freely generated by 0, +1;; ⊦ i + j = i + i0 ↔ j = i0; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ m +1 = m + 1; (s) ⊦ m + n = m + n0 ↔ n = n0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof Tree for nat2int-05 (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree